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JEE • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 211-240 of 264 questions

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211 Easy
211 of 264
Understanding +4 | -1
Which of the following cannot show variable oxidation state?
View answer and explanation
Correct option: B
Fluorine
Option B (Fluorine) is correct because fluorine is the most electronegative element and does not have d-orbitals available for expansion of its octet, thus it only exhibits a -1 oxidation state.
212 Medium
212 of 264
Application +4 | -1
How many electrons are involved in oxidation of KMnO4\mathrm{KMn}{\mathrm{O}_4} in basic medium?
View answer and explanation
Correct option: A
1
Option A (1) is correct because in a basic medium, KMnO4 is reduced to K2MnO4, involving a change of +7 to +6 in oxidation state, which means only 1 electron is involved.
213 Medium
213 of 264
Application +4 | -1
The oxidation number of ‘ P ’ in Mg2P2O7M{g_2}{P_2}{O_7}^{^{}} is
View answer and explanation
Correct option: B
+5+ 5
The correct answer is +5 because the oxidation number of P in Mg2P2O7Mg_2P_2O_7 is calculated based on the known oxidation numbers of Mg and O.
214 Medium
214 of 264
Application +4 | -1
In which of the following compounds iron has lowest oxidation number?
View answer and explanation
Correct option: C
Fe2O\mathrm{Fe}_2 \mathrm{O}
Option C (Fe2O\mathrm{Fe}_2 \mathrm{O}) is correct because the oxidation state of iron in this compound is +1, which is the lowest among the given options.
215 Medium
215 of 264
Analyzing +4 | -1
Which of the following is disproportionation reaction?
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Correct option: D
2NO2+2OH−→NO2−+NO3−+H2O2 \mathrm{NO}_2+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_2^{-}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O}
Option D (2NO2+2OH−→NO2−+NO3−+H2O2 \mathrm{NO}_2+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_2^{-}+\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{O}) is correct because nitrogen in NO2NO_2 is both oxidized (to NO3−NO_3^-) and reduced (to NO2−NO_2^-).
216 Hard
216 of 264
Analyzing +4 | -1
Assertion :- Fe3+\mathrm{Fe}^{3+} catalysts oxidation of I−\mathrm{I}^{-} by S2O82−\mathrm{S}_2 \mathrm{O}_8^{2-}

Reason :- S2O82−\mathrm{S}_2 \mathrm{O}_8^{2-} oxidize Fe2+\mathrm{Fe}^{2+} to Fe3+\mathrm{Fe}^{3+}
View answer and explanation
Correct option: C
If Assertion is True but the Reason is False
Option C is correct because while the assertion is true, the reason is false; Fe3+ catalyzes the reaction, but S2O82- oxidizes Fe2+ to Fe3+ not the other way around.
217 Medium
217 of 264
Understanding +4 | -1
What would happen when a small quantity of H2O2{\mathrm{H}_2}{\mathrm{O}_2} is added to a solution of FeSO4\mathrm{FeS}{\mathrm{O}_4}?
View answer and explanation
Correct option: D
An electron is lost by Fe2+\mathrm{F}{\mathrm{e}^{2 + }}
Option D (An electron is lost by Fe2+\mathrm{F}{\mathrm{e}^{2 + }}) is correct because H2O2{\mathrm{H}_2}{\mathrm{O}_2} acts as an oxidizing agent, oxidizing Fe2+\mathrm{F}{\mathrm{e}^{2 + }} to Fe3+\mathrm{F}{\mathrm{e}^{3 + }}.
218 Easy
218 of 264
Analyzing +4 | -1
In which reaction change in oxidation number of underlined atom is maximum when all of them react with Cl2C{l_2} water:-
View answer and explanation
Correct option: D
I2‾\underline {{I_2}}
Option D (I2‾\underline {{I_2}}) is correct because the oxidation number of iodine changes from 0 to +5, resulting in the maximum change.
219 Medium
219 of 264
Remembering +4 | -1
Which of the following reactions leads to the breakdown of a compound into two or more components?
View answer and explanation
Correct option: C
Decomposition reactions
Option C is correct because decomposition reactions involve the breakdown of a compound into two or more components.
220 Easy
220 of 264
Application +4 | -1
In Which of the following compounds the oxidation number of carbon is maximum:
View answer and explanation
Correct option: B
CHCl3\mathrm{CHCl}_3
Option B (CHCl3\mathrm{CHCl}_3) is correct because the oxidation number of carbon in this compound is +2, which is the highest among the given options.
221 Easy
221 of 264
Application +4 | -1
Oxidation number of oxygen in O2F2{O_2}{F_2} is
View answer and explanation
Correct option: C
+1+1
Option C (+1) is correct because fluorine is more electronegative than oxygen, assigning it a -1 oxidation state, leading to +1 for oxygen.
222 Medium
222 of 264
Analyzing +4 | -1
Assertion (A): Addition of water to a solution containing solute and solvent cannot change it's normality or molarity.

Reason (R): The milliequivalent and millimoles of the solute are changed on dilution.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D (Both (A) and (R) are false) is correct because adding water changes normality and molarity, and the milliequivalents and millimoles of solute remain constant upon dilution.
223 Medium
223 of 264
Application +4 | -1
MnO4−\mathrm{MnO}_4^ - is a good oxidising agent in different medium changing to MnO4−→Mn2+\mathrm{MnO}_4^ - \to \mathrm{M}{\mathrm{n}^{2 + }},→MnO42−,→MnO2,→Mn2O3\to \mathrm{MnO}_4^{2 - }, \to \mathrm{Mn}{\mathrm{O}_2}, \to \mathrm{M}{\mathrm{n}_2}{\mathrm{O}_3} Changes in oxidation number respectively are
View answer and explanation
Correct option: C
5,1,3,4
Option C (5,1,3,4) is correct because it accurately reflects the changes in oxidation number of Mn in the given transformations.
224 Easy
224 of 264
Analyzing +4 | -1
In C+H2O→CO+H2;H2Oactsas:\mathrm{C} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{CO} + {\mathrm{H}_2};{\mathrm{H}_2}\mathrm{O acts as}:
View answer and explanation
Correct option: A
Oxidant
Option A (Oxidant) is correct because in the reaction C+H2O→CO+H2\mathrm{C} + {\mathrm{H}_2}\mathrm{O} \to \mathrm{CO} + {\mathrm{H}_2}, hydrogen in water gains electrons (reduction), so water acts as an oxidizing agent (oxidant).
225 Medium
225 of 264
Analyzing +4 | -1
The redox reaction among the following is
View answer and explanation
Correct option: C
combination of dinitrogen with dioxygen at 200 K
The correct answer is C (combination of dinitrogen with dioxygen at 200 K) because the oxidation states of nitrogen and oxygen change during the reaction.
226 Easy
226 of 264
Understanding +4 | -1
An element with electronic arrangement at 2, 8, 18, 1 will exhibit the following oxidation states
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Correct option: B
12
Option A is correct because an element with the electronic configuration 2, 8, 18, 1 (Copper) can exhibit multiple oxidation states due to the involvement of d-electrons.
227 Easy
227 of 264
Application +4 | -1
In the reaction, CH3OH→HCOOH\mathrm{C}{\mathrm{H}_3}\mathrm{OH} \to \mathrm{HCOOH}, the number of electrons that must be added to the right is:
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Correct option: A
4
Option A (4) is correct because the oxidation state of carbon changes from -2 to +2, requiring 4 electrons to be added to the right side of the equation.
228 Easy
228 of 264
Remembering +4 | -1
A group of methods of quantitative chemical analysis involving the measurement of volume of reacting substance is known as :
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Correct option: B
Volumetric analysis
Option B (Volumetric analysis) is correct because volumetric analysis involves measuring the volume of reacting substances to determine the quantity of an analyte.
229 Easy
229 of 264
Application +4 | -1
The oxidation number of sulphur in H2 S2O7\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 and iron in K4Fe(CN)6\mathrm{K}_4 \mathrm{Fe}(\mathrm{CN})_6 is respectively:
View answer and explanation
Correct option: A
+6+6 and +2+2
Option A (+6 and +2) is correct because the oxidation number of sulfur in H2S2O7 is +6 and the oxidation number of iron in K4Fe(CN)6 is +2.
230 Medium
230 of 264
Analyzing +4 | -1
Assertion (A): Oxygen has oxidation state of -2 in both O2\mathrm{O}_2 and O3.\mathrm{O}_3.

Reason (R): Oxygen is assigned an oxidation state of -2 in almost all its compounds.
View answer and explanation
Correct option: D
Both (A) and (R) are false
Option D is correct because the oxidation state of oxygen in O2O_2 is 0, in O3O_3 is 0, and the general rule of -2 has exceptions like peroxides and superoxides.
231 Medium
231 of 264
Application +4 | -1
Among NH3,HNO3,NaN3\mathrm{N}{\mathrm{H}_3},\mathrm{HN}{\mathrm{O}_{3,}}\mathrm{Na}{\mathrm{N}_3} and Mg3N2;\mathrm{M}{\mathrm{g}_3}{\mathrm{N}_2}; the number of molecules having nitrogen in negative oxidation state is
View answer and explanation
Correct option: C
3
Option C (3) is correct because NH3, NaN3 and Mg3N2 have nitrogen in negative oxidation state.
232 Medium
232 of 264
Analyzing +4 | -1
Assertion: In alkaline medium, KMnO4\mathrm{KMnO}_4 acts as powerful oxidising agent.
Reason: KMnO4\mathrm{KMnO}_4 reduces to give Mn2+\mathrm{Mn}^{2+} in alkaline medium.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect
The assertion is correct; KMnO4 is a powerful oxidizing agent in alkaline medium. However, the reason is incorrect; KMnO4 reduces to Mn+6 in alkaline medium, not Mn+2.
233 Medium
233 of 264
Analyzing +4 | -1
Identify the disproportionation reaction :
View answer and explanation
Correct option: D
2NO2+2OH−→NO2−+NO3−+H2O2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O
Option D (2NO2+2OH−→NO2−+NO3−+H2O2N{O_2} + 2O{H^ - } \to NO_2^ - + NO_3^ - + {H_2}O) is correct because nitrogen undergoes both oxidation and reduction in the same reaction.
234 Easy
234 of 264
Analyzing +4 | -1
Which of the following is not a redox reaction?
View answer and explanation
Correct option: D
NaCl+H2SO4→NaHSO4+HCl\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCl}
Option D is correct because the reaction NaCl+H2SO4→NaHSO4+HCl\mathrm{NaCl}+\mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{NaHSO}_4+\mathrm{HCl} does not involve any change in the oxidation state of any element.
235 Medium
235 of 264
Application +4 | -1
Which of the following is the cell reaction that occurs when the following half - cells are combined?

I2+2e−→2I−(1 M);E∘=+0.54 VBr2+2e−→2Br−(1 M);E∘=+1.09 V\begin{aligned} & \mathrm{I}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{I}^{-}(1\,\mathrm{M}) ; \mathrm{E}^{\circ}=+0.54 \mathrm{~V} \\ & \mathrm{Br}_2+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(1\,\mathrm{M}) ; \mathrm{E}^{\circ}=+1.09 \mathrm{~V} \end{aligned}
View answer and explanation
Correct option: C
2I−+Br2→I2+2Br−2 \mathrm{I}^{-}+\mathrm{Br}_2 \rightarrow \mathrm{I}_2+2 \mathrm{Br}^{-}
Option C (2I−+Br2→I2+2Br−2 \mathrm{I}^{-}+\mathrm{Br}_2 \rightarrow \mathrm{I}_2+2 \mathrm{Br}^{-}) is correct because Bromine has a higher reduction potential and will be reduced, while Iodide will be oxidized.
236 Medium
236 of 264
Application +4 | -1
The value of 'n' in the reaction Cr2O72−+14H++nFe2+→2Cr3++nFe3++7H2O\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 14{\mathrm{H}^ + } + n\mathrm{F}{\mathrm{e}^{2 + }} \to 2\mathrm{C}{\mathrm{r}^{3 + }} + n\mathrm{F}{\mathrm{e}^{3 + }} + 7{\mathrm{H}^2}\mathrm{O} will be
View answer and explanation
Correct option: C
6
Option C (6) is correct because balancing the redox reaction shows that 6 moles of Fe2+ are required to react with 1 mole of dichromate.
237 Medium
237 of 264
Remembering +4 | -1
Starch iodide paper is used to test for the presence of:
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Correct option: A
Iodine
Option A (Iodine) is correct because starch iodide paper turns blue in the presence of iodine.
238 Medium
238 of 264
Understanding +4 | -1
In which case +1+ 1 oxidation state is stable than +3+ 3?
View answer and explanation
Correct option: C
Tl\mathrm{Tl}
Option C (Tl\mathrm{Tl}) is correct because Thallium (Tl) exhibits a stable +1 oxidation state due to the inert pair effect, which makes the +3 oxidation state less stable.
239 Easy
239 of 264
Understanding +4 | -1
The halogen that most easily reduced is
View answer and explanation
Correct option: A
F2{F_2}
Option A (F2{F_2}) is correct because fluorine is the most electronegative element, making it the strongest oxidizing agent and thus the most easily reduced halogen.
240 Medium
240 of 264
Understanding +4 | -1
It is found that V forms a double salt isomorphous with Mohr’s salt. The oxidation number of V in this compound is :
View answer and explanation
Correct option: B
+2+ \mathrm{ }2
Option B (+2+ \mathrm{ }2) is correct because Mohr's salt has the formula FeSO4.(NH4)2SO4.6H2O\mathrm{FeS}{\mathrm{O}_4}\mathrm{.}{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{S}{\mathrm{O}_4}.6{\mathrm{H}_2}\mathrm{O}, and if V forms a similar double salt, it replaces Fe with the same oxidation state of +2.
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